Question: In the diagram, $BP$ and $BQ$ trisect $\angle ABC$. $BM$ bisects $\angle PBQ$. Find the ratio of the measure of $\angle MBQ$ to the measure of $\angle ABQ$.
Let $\angle MBQ = x$, so $\angle MBP=x$ as well.  Therefore, we have $\angle PBQ = 2x$, so $\angle ABP = \angle PBQ = \angle QBC = 2x$.  Finally, we have $\angle ABQ = \angle ABP + \angle PBQ = 4x$, so  \[\frac{\angle MBQ}{\angle ABQ} = \frac{x}{4x} = \boxed{\frac14}.\]